The order of a group is its cardinality, i.e., the number of its elements;
the order, sometimes period, of an element a of a group is the smallest positive integerm such that am = e (where e denotes the identity element of the group, and am denotes the product of m copies of a). If no such m exists, we say that a has infinite order. All elements of finite groups have finite order.

Question: If I have where , is it correct to say that . If so, then contains elements. Must all the elements be distinct? Must an identity be in the group?

If you mean that G is a cyclic group (in other words, the entire group is ) such that is as high as you can get without repeating, then actually because has 5 elements (count them).

Yes, by definition, a group must have an identity, which in this case is , and all elements must be distinct. For example, although and look different, they're actually the same element because in a cyclic group, the elements start ``repeating.''

Aug 15th 2010, 12:35 PM

novice

Since and , is it of the order of or ?

Drawing from what is said in Wikipedia, when is the smallest positive integer such that , then must be the number, then what is the order? hmm?

Aug 15th 2010, 12:51 PM

WolfTecc

In my example, since , the order of the group is 5. But what exactly do you mean when you say, "I have where "? We can raise to any power we want and the result will always be in the group . However, if this is a finite group, we'll get a lot of repetition.

Let's say you start out with , raise it to some power , and get the same thing as the identity, . If is the smallest positive integer power such that , then is the order of the group. In my example, since 5 is the smallest power that gives me back the identity, that's the order of the group.

Thus, if I have a group of order , since I don't want duplicates in my listing, I'll only raise to the powers . This will give me every unique element in the group.

It shows a symmetric group table. In the example, each element , and square to . Then it says that these group elements have order 2.

I am a little confused by the terms "order of a group" and "order of group elements." Are they the same?

Aug 15th 2010, 01:27 PM

WolfTecc

Nope, they're not the same--it's confusing at first, I know. The ``order of a group'' G is the number of elements it contains. Consider The order of this group is 4 since it has 4 elements.

The ``order of an element'' means: to what smallest power do I have to raise this element to get back the identity? Consider the element in the group I described above. I know. Thus, the element has order 2. So you see that the order of a group and the order of an element are not the same, but in fact, the order of an element always divides the order of a group (notice that 2, the order of , divides 4, the order of the group.)

In your example from Wikipedia, it says if you square those elements, you get the identity. By definition, then, these elements have order 2. Make sense?

Aug 15th 2010, 06:46 PM

novice

Since , and square into , which means that , similarly and . If , then it's trivial. Now, if , how do you explain ? What is the simplest example?

Aug 15th 2010, 08:39 PM

WolfTecc

So we're assuming that which implies (you can see this by multiplying each side of the first equation by ). Thus, you cannot have both and --it's a contradiction.

If an element has order 2, it is its own inverse, i.e. . Certainly, if , then . But doesn't have to be the identity for this to happen.

Let . Then consider what happens when we square . We'll have . So we know in this group, following similar reasoning to the first paragraph.

So in summary, if squares to , you definitely know that But doesn't necessarily have to be the identity.

Aug 15th 2010, 09:16 PM

novice

In terms of a real number, if .

What will be where , and , ? I can't think of any. Can it be an imaginary number?

Aug 15th 2010, 09:46 PM

WolfTecc

If "Multiplication" is the group operation, then we're treating 1 as the identity for this group. Keep in mind, this means 0 is not the identity for this group. (0 is the identity for a different group, where "addition" is the operation)

What about ? We know , regardless of what is. Since , from previous discussion we know . So satisfies the criteria, doesn't it?

In fact must be a real number since when you square a complex number, you double its angle (measured counter-clockwise about the origin from the real axis). So the only possibility is for to have angle 180 (so that it's a negative number) or angle 0 (so that it's a positive number).

Aug 16th 2010, 05:12 AM

novice

Sir, I want to thank you for your time for answering all my questions. You gave very clear instructions at every step.